Problem: An object is launched from a platform. Its height (in meters), $x$ seconds after the launch, is modeled by: $h(x)=-5(x-4)^2+180$ How many seconds after being launched will the object hit the ground?
Answer: The object hits the ground when $h(x)=0$. $\begin{aligned} h(x)&=0 \\\\ -5(x-4)^2+180&=0 \\\\ -5(x-4)^2&=-180 \\\\ (x-4)^2&=36 \\\\ \sqrt{(x-4)^2}&=\sqrt{36} \\\\ x-4&=\pm6 \\\\ x&=\pm6+4 \\\\ x=10&\text{ or }x=-2 \end{aligned}$ We found that $h(x)=0$ for $x=10$ or $x=-2$. Since $x=-2$ doesn't make sense in our context, the only reasonable answer is $x=10$. In conclusion, the object will hit the ground after $10$ seconds.